In the previous lecture, we had defined quantities like the first moment of an area and centroid mathematically and said that these are related to problems in mechanics. In this lecture, we are going to solve some problems using these concepts. As I had indicated towards the end of my previous lecture on this topic, the, the place where we use these concepts is where we have a distributed force. What do we mean by that? For example, if I have a beam and there is some mass on top of it so that it applies a force on the beam, the force maybe described by a function f(x) so that if I take a small section here delta x length, the force on this section, delta F is equal to f(x) times delta x. So f(x) is nothing but force per unit length. It is in dealing with such distributed forces that the concepts developed in the previous lecture are going to be handy So the question we ask is given this distribution of force f(x) what is the total force on the system and where effectively is it acting? Let me explain that a little further. The total force is going to be summation of that delta F that is acting on a small section of length delta x and this is going to be summation f(x) delta x which in the limit is going to be integration f(x) dx, that is the net force And when we say where effectively is it acting, that means what moment or torque should I apply to this beam in order to keep it in equilibrium. For example, if I have this beam fixed at this point or let me put a pin joint here, what torque should I apply here in order that this beam is in equilibrium or equivalently at which point should I apply this net force F that I have calculated above here so that the effect of this force both the torque as well as the net force is nullified? To do that, one we require that summation FY where Y is this direction be 0 and that gives me the net force F should be equal to f(x) dx. The 2nd equilibrium condition is that the torque about this pin joint vanish and that requires that the distance of the force that I am applying, call this X be such that it nullifies the torque generated by this force f(x). So F times X should be equal to summation x delta F which is nothing but integration f(x) x dx So on this beam if there is a force distribution f(x) then I have the net force F that I am supposed to apply this is about this pin joint to equilibrate the beam is going to be f(x) dx and I should have F times X where X is the distance at which the force is being applied

equal to integration x f(x) dx which is nothing but the moment generated by the force distribution f(x). And therefore X equals integration x f(x) dx over F And this by definition is the definition of centroid. Therefore, the net force is the area of this force distribution curve and the point at which effectively this force acts is the centroid of this area formed by the beam and this force distribution curve That is how we use the concept of the first moment or the centroid. I must point out that when the total force capital F is applied at the centroid, no other force is needed to support the beam That is, in that situation, the force applied by the pin joint will be 0. As such, in the case of f(x) being the gravitational force, the centroid gives the position of the center of gravity. Let us take some examples Suppose I have a beam. Let us call this point x equal to 0 of length L and there is a force distribution of the form of a rectangle from point x1 to x2. In that case, one can easily see, suppose this magnitude is W, one can easily see that the net force that this applies is the area of this rectangle which is going to be x2 minus x1 times W. And where does it act? It acts at the centroid of the area formed by this force distribution and the beam And x centroid is nothing but x1 plus x2 divided by 2. Therefore I can replace this entire force or represent this entire force like this. This is a beam, this is where it is hinged. The net force is of the amount W x2 minus x1 acting at a distance this is x1, this is x2 at a distance let me write it with blue, this is x1 plus x2 divided by 2. So that is one example Next, we consider triangular loading. In that I have a beam of some length L and it is loaded like this where the maximum load is per unit length is W. Load starts at a distance x1 and goes all the way up to x2 and what we want to figure out is how much is the total force acting on the beam and where effectively is it acting? So total force is going to be the area of this triangle which is W x2 minus x1 divided by 2 and where it acts is at the centroid of this triangle Recall from the previous lecture that if I am given a triangle then with respect to one of these corners, if this distance is a and this is b, then the centroid xc is given as one third a plus b. In the present case, the 2 points are at a distance of x2 minus x1 Both the points are at a distance of x2 minus x1 from this corner. And therefore, the centroid xc is going to be at a distance from this point the hinge point here x1 plus two thirds

x2 minus x1 or this comes out to be x1 over 3 plus 2 x2 over 3 is equal to one third x1 plus 2 x2 So if I were to look at this load effectively how it is working if this is the beam then the load can be effectively replaced by a force of W x2 minus x1 divided by 2 acting at a distance of one third x1 plus 2 x2 from this point. This is another example of how we apply the concept of first moment and the centroid in mechanics Next, I will consider one more example where the loading is trapezoidal. That means, the loading starts at x1 but it has a finite amount w1. Then it goes up to w2 per unit length and distance x2. Along the way this is how the load is distributed. Again, with respect to this point where the beam is hinged, I want to find out what is the total load and where effectively is it acting? So the total load is going to be the area of this trapezoid I can for later references divide this area into 2 areas, one which is corresponding to this rectangle of height w1 and width x2 minus x1 and another one, the triangle So the total area is going to be the area of the rectangle which is w1 x2 minus x1 plus the area of the triangle which is going to be w2 minus w1 divided by 2 times x2 minus x1 which comes out to be the area of trapezoid which is nothing but one half w1 plus w2 times x2 minus x1. So this is the net load which is working on this. To calculate where it acts, I am going to use a… an observation that we made last time I know that this rectangle where load acts right in the middle at a distance of x2 plus x1 divided by 2 from this point. Similarly the triangular loading which we just calculated in the previous slide acts at x1 plus 2 x2 divided by 3 distance from this point. So I can take these 2 loads and then calculate what will be the effective centroid for this entire area and that we know from our previous lecture is going to be xc equals area of the rectangle times xc of the rectangle, just making it symbolically, plus area of the triangle times xc of the triangle divided by the total area. Total area which is the net force we have already calculated. We also know the positions of the centroid of the rectangle. We also know the position of the centroid of the triangle and we know the area for both. And therefore we can calculate xc. You do a quick calculation and the answer you get is xc equals x1 plus x2 times w1 plus one third x1 plus 2 x2 w2 minus w1 divided by w1 plus w2 Let me write it again. So what we are considering is this beam which is loaded from point x1 to point x2 with this trapezoidal load. Then the total

load force it comes out to be one half w1 plus w2 times x2 minus x1 and the point at which it acts is at a distance xc which is equal to, as I wrote in the previous slide, x1 plus x2 times w1 plus one third x1 plus 2 x2 times w2 minus w1 divided by w1 plus w2 If I take x1 to be 0 that is the load starts right here and goes up to x2 then the centroid comes out to be x2 w1 plus two thirds x2 w2 minus w1 divided by w1 plus w2, which is nothing but x2 over 3 2 w2 plus w1 divided by w1 plus w2. So that is the example for trapezoidal loading Next, let us now solve a problem with a particular loading. So suppose I have a beam of length l. It is on a roller on this side and on a pin joint on this side and it is loaded with triangular loading like this with these being W0, W0. This length being 2 l by 3 and this length therefore being l by 3. And I wish to calculate the reactions at the pin joint and at the roller. So for that, what I need to do is take this beam and make a free body diagram So we will assume that reaction at this point is N2, reaction at this point is N1. And the 2 forces due to the loading, okay the forces due to the loading I will split them into 2, 1 due to this triangle to the left and one due to this triangle to the right. I will take the triangle on the left acting like this at a distance which we have to calculate And similarly, the triangle to the right also applies a load, let us call this F1 and F2 at a distance from this point Now what we did earlier I can take F1 to be acting at centroid of this 1st triangle. F1 is going to be equal to one half W0 times 2 l by 3. So this is going to be W nought l by 3. And where does it act? It acts at the centroid of this triangle. The centroid of this triangle xc is going to be equal to one third, a in this case is 0, this is the point where a is and b is 2 l divided by 3 So this is acting at 2 l by 9. So this distance is 2 l by 9 and F1 is equal to W nought l by 3. Similarly for the other triangle, the force F2 is going to be the area of that triangle which is going to be one half W nought l by 3 which is W nought l by 6. So this force

is equal to W naught l by 6. And it acts at the centroid of the right hand triangle The centroid with respect to this point, the point in the middle is going to be at a distance of one third of l by 3 plus l by 3 which is one third of 2 l by 3, 2 l by 9. So the distance from this point, the corner is going to be 2 l by 3 plus 2 l by 9 which is nothing but 8 l by 9. So the force here acts at 8 l by 9. We are now ready to solve the problem So what we have done effectively is took this beam which is loaded like this with this being W0, this distance being 2 l by 3, this distance being l by 3. Then we said, the reactions at these points are going to be N1 and N2 And the 2 triangles we replace by one force acting downwards which is W nought l by 3 acting at a distance of 2 l by 9 And the other force acting here whose magnitude is given by W nought l by 6 and it is acting at a distance of 8 l by 9. So I can forget about these triangles now and instead focus on these forces and do my calculations as we have been doing in the beginning of the 1st few lectures of this course A beam is loaded with these 2 forces and there are these are 2 reactions, simple N1 plus N2 being equal to W nought l by 3 plus W nought l by 6 which comes out to be W nought l by 2. And then we balance the moments about this point. All the distances are known and when we solv…. so it’s going to be N2 times l is going to be equal to 2 l by 9 times W nought l by 3 plus 8 l by 9 times W nought l by 6 When I solve the 2 equations, I get N2 to be equal to two ninths W nought l and N1 to be equal to 5 W nought l over 18. So that is how we have used the concept of centroid in finding out where effectively the force given by a distribution acts, what is its moment and then applying a regular point force diagrams to calculate the reaction forces and so on